Question

Need help with the following problem: only one point (x,y)

The next two parts deal with the potential away from the straight line joining the two electrodes. We will be looking for points that have the same potential as (4,4) cm, i.e., we want to map the equipotential line. Great! You have located the point along y = 6 cm that has the same potential as (4,4) cm. To further map the equipotential line, where along the y = 7 cm is the potential the same as (4,4) cm? Click on the graph, then click on the "Submit Answer" button. To find the location where it has the same potential as (9,4) cm, the round electrode is first approximated as point change. This means that the potential at a distance r from each charge is v = kq/r. Since the charges do not change, the equipotential line is determined by points which have the same ration in the distances between the two electrodes. Now, since the ratio of the distance from the positive electrode to the reference point over the distance from negative electrode is 1/3, all the points along the equipotential line have the ratio in the distances.

Answer 1

(1, 7)

We require the point on the line y=7, such that the ratio of its distance from the positive charge to its distance from the negative charge is 1/3.

We require the point on the line y = 7, such that the ratio of its distance from the positive charge to its distance from the negative charge is 1/3. Therefore r_p/r_n = 1/3 Let the point be at a distance x in the x-direction from the positive charge Therefore r_p/r_n = squareroot 3^2 + x^2/squareroot 3^2 + (8 - x)^2 = 1/3 Solving for x, we get, x = -1. This means that the point has the x-coordinate at a distance of 1 unit towards the left of the positive charge. Therefore, the co-ordinates of the equipotential point along y = 7 are: (1, 7)

Let the point be at a distance x in the x-direction from the positive charge.

$\therefore\ \frac{r_p}{r_n}\ =\ \frac{\sqrt{3^2\ +\ x^2}}{\sqrt{3^2\ +\ (8-x)^2}}\ =\ \frac{1}{3}$

Solving for x, we get,

x = -1.

This means that the point has the x-coordinate at a distance of 1 unit towards the left of the positive charge.

Therefore, the co-ordinates of the equipotential point along y=7 are:

(1, 7)